# Let $f(x) =x^2$ and $g(x)=\sin x$ for all $x\in R$ .Then the set of all $x$ satisfying $(fogogof)(x)=(gogof)(x)$ where $fog(x)=f(g(x))$ is

$\begin{array}{1 1}(a)\;\pm \sqrt{n\pi};n\in \{0,1,2\}\\(b)\;\pm\sqrt{ n\pi};n \in \{1,2\}\\(c)\;\large\frac{\pi}{2}+\normalsize 2n\pi;n\in \{...-2,-1,0,1,2\}\\(d)\;2n\pi;n\in \{...-2,-1,0,1,2...\}\end{array}$

Given :
$f(x) =x^2$
$g(x)=\sin x$
$(gof)(x)=g(f(x))=g(x^2)$
$\Rightarrow \sin x^2$
$go(gof)(x)=\sin(\sin x^2)$------(1)
$(fogogof)(x)=(\sin(\sin x^2))^2$-----(2)
Step 2:
Given :
$(fogogof)(x)=(gogof)(x)$
$(\sin (\sin x^2))^2=\sin(\sin x^2)$
$\sin(\sin x^2)\{\sin(\sin x^2)-1\}=0$
$\sin(\sin x^2)=0$
$\Rightarrow \sin x^2=0$
$x^2=n\pi$
$x=\pm \sqrt{n\pi}$
Where $n\in \{1,2....\}$
Hence (b) is the correct answer.